Convergence of Iterative Voting

AAMAS 2012Valencia, Spain

Omer Lev & Jeffrey S. Rosenschein

What is Iterative Voting?Color of the new car…

Adam:

Eve:

Cain:

Abel:

Seth:

(Seth breaks ties)

What is Iterative Voting?Color of the new car…

Adam:

Eve:

Cain:

Abel:

Seth:

Wait a minute!

What is Iterative Voting?Color of the new car…

Adam:

Eve:

Cain:

Abel:

Seth:

What is Iterative Voting?Color of the new car…

Adam:

Eve:

Cain:

Abel:

Seth:

Wait a minute!

What is Iterative Voting?Color of the new car…

Adam:

Eve:

Cain:

Abel:

Seth:

What is Iterative Voting?Color of the new car…

Adam:

Eve:

Cain:

Abel:

Seth:

Can’t we all just get along?

What we know:(Meir et al. – AAAI

2010)Assuming players play a myopic “best response” – reacting to the current state:

2 cases: Randomized tie breaking rule:

from truthful state Deterministic tie breaking

rules: from any state (including non-truthful)

Iterative Plurality converges

And linear ordered – i.e., there is a fixed order between candidates, according to which ties are resolved

Tie-breaking rulesLinear: ≻ ≻≻ ≻

Non-linear:

There is no set order between red and orange

Pastry example:(thanks to Ilan Nehama)

Short aside:What are scoring

rulesScoring rules for m candidates define a scoring vector:

under the condition

A voter gives α1 points to his most preferred candidate, α2 points to his 2nd preference, etc.

The winner is the candidate with most points

Short aside:Examples of

scoring rulesPlurality: (1,0,…,0,0)

Veto: (1,1,…,1,0)

Borda: (m-1,m-2,…,1,0)

k-approval: (1,1,…,1,0,0,…,0)k candidates

k-veto: (1,1,…,1,0,0,…,0)k candidates

Theorem I: Tie-breaking rules

matterWhen using any arbitrary

tie-breaking rule (i.e., not necessarily linear

ones), every scoring rule & Maximin has tie-

breaking rule for which it will not always converge

Theorem I: Proof sketch  (scoring rules)4 candidates, 2 voters, tie breaking rule makes c win if not tied with b. b wins if not tied with d. d wins if not tied with a.a … b c d≻ ≻ ≻ ≻

c … d b a≻ ≻ ≻ ≻b … a d c≻ ≻ ≻ ≻c … d b a≻ ≻ ≻ ≻

b … a d c≻ ≻ ≻ ≻d … c a b≻ ≻ ≻ ≻

a … b c d≻ ≻ ≻ ≻d … c a b≻ ≻ ≻ ≻

Theorem II: Borda doesn’t work

When using the Borda voting rule, regardless of

tie-breaking rules, the iterative process may

never converge

Theorem II: Proof sketch4 candidates, 2 voters (tie breaking doesn’t matter):

a b c ≻ ≻ d≻

c d b ≻ ≻ a≻

d – 2; a, b – 3; c – 4

b a d ≻ ≻ ≻cc d b ≻ ≻ ≻aa – 2; c, d – 3; b – 4

b a d ≻ ≻ ≻cd c a ≻ ≻ ≻bc – 2; a, b – 3; d – 4

a b c ≻ ≻ d≻

d c a ≻ ≻ b≻

b – 2; c, d – 3; a – 4

Theorem III: Iterative Veto

converges

When using linear tie-breaking rules, iterative Veto will always converge

– from truthful or non-truthful starting point

Theorem III: Proof“Best response” straight-forwardly defined as vetoing the current (unwanted) winner.

Lemma 1: If there is a cycle, taking a stage in the cycle where there is more than one candidate with the maximal score, suppose winner score is s. Then winning score at any other stage is s or s+1. Any stage with s+1 score has only one candidate with that score.

Theorem III: Proof Lemma 1

The futility of having a single winner – the score can’t get higher, and you can’t get multiple candidates to share the score:

ss+1

s-1 ss+1

s-1

Theorem III: ProofLemma 2: If there is a cycle, all stages with more than one candidate with the maximal score have the same number of candidates with maximal score and maximal-1 score, and these are the same candidates in all the cycle.

ss+1

s-1

Theorem III: Proof2 types of player moves:

A candidate with a score of s becomes winner with score of s+1

A candidate with a score of s-1 gets point and becomes winner

Previously vetoed candidates become winners (gaining a point), i.e., voters’ situation progressively

worse. This is a finite process

Theorem IV: k-Approval doesn’t

work

When using k-approval voting rule for k≥2, even with linear tie-breaking

rule, the iterative process may never converge

Theorem IV: Proof sketch4 candidates, 2 voters, and the tie breaking rule is alphabetical (a b ≻

c d)≻ ≻b d c ≻ ≻

a≻a d c ≻ ≻

b≻d – 2; a, b – 1; c – 0

b d c ≻ ≻ a≻

a c d ≻ ≻ b≻

a, b, c, d – 1

b c d ≻ ≻ a≻

a c d ≻ ≻ b≻

c – 2; a, b – 1; d – 0

b c d ≻ ≻ a≻

a d c ≻ ≻ b≻

a, b, c, d – 1

Future workBetter understanding of what influences convergence (tie-breaking rules identified, what else?)

What is best-response for complex voting rules?

Weighted games

Computational complexity issues for best-response in complex voting rules

Moving beyond myopic best-response to more complex and varied responses

Fin

Thanks for listening!

(guess they decided to compromise on the car colors…)